ΔH = -1031 Kj/mol
Propanol 2C3H7OH + 9O2 → 6CO2 + 8H2O
Bond Breaking Bond Making
2 x 7(C-H) = 14 x 412 6 x 2(C=O) = 12 x 743
2 x 2(C-C) = 4 x 348 8 x 2(O-H) = 16 x 463
2 x (C-O) = 2 x 360 = -16324
2 x (O-H) = 2 x 463
9 x (O=O) = 9 x 496
= 13270
ΔH = -3054/2
ΔH = -1527 Kj/mol
Butanol C4H9OH + 6O2 → 4CO2 + 5H2O
Bond Breaking Bond Making
9(C-H) = 9 x 412 4 x 2(C=O) = 8 x 743
3(C-C) = 3 x 348 5 x 2(O-H) = 10 x 463
(C-O) = 360 = -10574
(O-H) = 463
6 x (O=O) = _ 6 x 496
= 8551
ΔH = -2023 Kj/mol
Pentanol 2C5H11OH + 15O2 → 10CO2 + 12H2O
Bond Breaking Bond Making
2 x 11(C-H) = 22 x 412 10 x 2(C=O) = 20 x 743
2 x 4(C-C) = 8 x 348 12 x 2(O-H) = 24 x 463
2 x (C-O) = 2 x 360 = -25972
2 x (O-H) = 2 x 463
15 x (O=O) = 15 x 496
= 20934
ΔH = -5038/2.
ΔH = -2519 Kj/mol
This shows that the alcohol with the most Carbons which means it has the most bonds gives off the most amount of energy. This indeed supports my prediction.
Preliminary experiment:
I set up the experiment exactly as I stated in my method. I only did my preliminary with the alcohols of Methanol and Pentanol. These two are the liquids with the least and highest number of carbons respectively. This meant that I could investigate the boundaries of my experiment, and also to check whether my predictions were correct.
Results:
This all agrees with my prediction and also my quantative prediction. One problem however that I must change to overcome, is that my thermometer is at the bottom of the tin-therefore most probably only reading the bottom of the tins temperature-not the waters temperature. So to overcome this I will have to somehow make sure that the thermometer is in the middle of the water by sticking it to the side with tape.
OBSERVATION
During the experiment I noted that actually different alcohols had different flame sizes. It seemed that as the number of carbons in the alcohol increased the size flame increased. I stated previously that I would be keeping the distance between the burner wick and the bottom of the tin constant throughout. This I did for all alcohols apart from pentanol. Although pentanol is the largest alcohol I have here the flame size was much smaller than any other. This is because the alcohol is much more efficient that the others (more energy is released using less substance). So for pentanol I decreased the length between wick and
tin bottom relative to all the other flame sizes.
I found the total energy of the reaction:
E = mCΔt
E = total energy
m = mass of water (100g)
C = specific heat capacity of water (4.18)
Δt = the change in temperature
Here you can see I took two readings of every Alcohol, taking one repeat. I recorded every reading, I made sure to include everything, even if obscure.
Luckily there were no readings that were harder to take than any others, the experiment was fairly consistent in its method and there was no need for sudden change in anything.
If you have a look at the difference between temperature changes of the same Alcohol then they seem a little far apart from each other. One must remember that for each Alcohol and each repeat different masses of Alcohol was burnt; this mass is proportional to the energy given off. Therefore if little energy was released then little mass was used, and vice versa. I carried on with this therefore finding out the results per mole next.
I think that the range of alcohols I used was good. I must take into account that using pentanol produced a large flame and the temperature rose up to at least 50°C. If I was to use the next alcohol in the chain (6) then its highly possibly that the water would actually boil. Thus water would be lost and I would not be able to measure the correct mass lost.
Next I found the energy released per mole by dividing the total energy by 1000n to give me an answer in Kj/mol
ΔH = the energy released per mole ΔH = E .
n = number of moles 1000n
E=energy released
I multiplied the n by 1000 to convert the answer from J to Kj.
Methanol:
Mr =32 change in mass =1.06 ÷ Mr 0.97
0.033125 0.0303125
x1000 x1000
=33.125 =30.3125
5040÷33.125 6720÷30.3125
=152.15 Kj/mol =221.69 Kj/mol
Ethanol:
Mr = 46 =3.84 =1.19
0.0834782609 =0.0258695652
x1000 x1000
83.478 25.869
15960÷83.478 7980÷25.869
=191.19 Kj/mol =308.48 Kj/mol
Propanol:
Mr = 60 =0.93 =0.90
0.0155 0.015
x1000 x1000
=15.5 =15
9240÷15.5 4200÷15
=596.13 Kj/mol =280 Kj/mol
Butanol:
Mr = 74 =1.59 =1.54
0.0214864865 0.0208108108
x1000 x1000
=21.49 =20.81
15540÷21.49 11340÷20.81
=723.13 Kj/mol =544.93 Kj/mol
Pentanol:
Mr = 88 =0.33 =0.63
0.00375 0.007159
x1000 x1000
- 7.16
4200÷3.75 4200÷7.16
=1120 Kj/mol =586.59 Kj/mol
ANALYSIS
As the results of the per mole calculations seemed so far apart (almost double/half of each other) I decided to take an average of the two readings to see if this gave me a clearer indication of the trend. The reason for the fluctuations in the results I will talk about later in the evaluation:
From this experiement I have found out that the enthalpy of an alcohol increases as its relative mass increases. This is shown in my graph below. Pentanol with a relative mass of 88 has the greatest enthalpy whereas Methanol has the lowest.
Graph:
Conclusion:
My conclusion agrees with my hypothesis which was “that as the relative mass of the alcohol increases (i.e. a greater number of carbons) the enthalpy will increase. Therefore Pentanol will release the greatest amount of energy and Methanol the least.
I predict that the smaller carbons will burn with a smaller flame thus giving out less energy. Therefore the liquids with the more numbers of carbons for example Butanol and Pentanol will burn with a much larger flame thus giving out more energy.”
I believe this is due to the way that the alcohols bond. They are a chain of carbons with a 0-H bond on the end.
1 Methanol (CH3OH)
2 Ethanol (C2H5OH)
3 Propanol (C3H7OH)
4 Butanol (C4H9OH)
5 Pentanol (C5H11OH) These all bond in the same manner.
Longer chain hydrocarbons give out more energy than shorter chain ones. The enthalpy increased as the relative mass of the alcohol increased. Methanol therefore gave out the least energy and Pentanol gave out the most. As I stated above all of the alcohols above bond in the same manner. Energy is needed to break the bonds and energy is released when new bonds are made. Looking at the above table I can see that Methanol has the least bonds and Pentanol has the most bonds. As Pentanol has the most bonds it is elementary to say that Pentanol needs more energy to break its bonds and more enery is given out when new bonds are made. It has the largest enthalpy as it gives off the most energy per mole, and methanol gives off the smallest amount of energy- therefore it has the smallest enthalpy.
I can see this too from my graph. My results followed my prediction. After drawing on by hand a line of best fit I can see that my results are actually approx. directly proportional to each other. They are not directly situated on the proportional line, but this can be expected as the results are averages. They still follow suit though. This further follows my prediction. I can safely conclude that the larger the relative mass of an alcohol, the great the enthalpy of it will be.
One must remember that my results were only an average of the two repeats. I took two readings for every alcohol. If you look back at page 6 you will see the singular results for each alcohol. I decided not to use these to draw my graph as they were too far apart from each other in repeats. And also in some cases between alcohols the results went down before going up as predicted. I worked out that this was due to an experimental error that I did during my observation. I will explain this in my evaluation. When I took an average of the two repeats for each alcohol I realized that they showed a much better trend and they supported my prediction much more solidly. This is why I chose to use the averages instead of the singular results.
Quantitative conclusion:
The enthalpies I calculated using data from my experiment should be similar to those I calculated using bond energy values.
Looking at the table I can see that the enthalpies calculated in my experiment were considerably smaller that those calculated using bond energies. There are however some similarities. These being that the distance between each result in one category stays approximately the same. Whereas in the other category the result is distance is different but still stays approx. the same throughout.
From this I can say that perhaps the energy released from the flame wasn’t necessarily going straight to heating the water. The energy could have been heating the tin and also being lost to the surrounding air in the 5cm distance between the wick and the bottom of the tin.
EVALUATION:
I felt the experiment and the observation went very well. Everything was easy to set up and then recording data was simple and after a few readings I got into a flow of data the results down in the correct time. There were no problems whilst taking the experiment.
There was however a problem with my repeat readings. They were extremely far apart from each other, in some cases one was half of the other. As almost each repeat was like that I decided to take an average to see whether this would show me a better trend. As seen on page 7 in the table these results showed a much better trend. I thought the average was much more reliable and showed the trend that should be expected for this experiment.
The fact that the results do not lie exactly on the line of best fit, which happens to be directly proportional is due to the averages being used. The only reason that averages were used is because the results were so far apart from each other. The reason for this is due to an experimental flaw in the way I carried out my procedure.
There are no drastically obscure results that can be seen on my graph (anomalies). They all lie slightly off the line of best fit. This is because of the fact that they are averages. When drawing the line of best fit I tried to draw it so that an equal number of points lay on either side of the line thus giving the best trend line possible.
Here I have compared my results to those I calculated using bond energy values:
From this I can say that perhaps the energy released from the flame wasn’t necessarily going straight to heating the water. The energy could have been heating the tin and also being lost to the surrounding air in the 5cm distance between the wick and the bottom of the tin.
Flaws in the experimental design-
The sizes of the different alcohol burners I was using varied; this meant that the size of the wick and therefore the flame was different. If I was to repeat the experiment I would try to make sure that all burners and wicks were the same size or as close as possible to reduce this undesired effect.
Lots of energy was being lost in the 5cm gap between the wick and the bottom of the tin. It was being lost to the surroundings as light and heat. I think this was not the best way to conduct the experiment. Looking back on the experiment again I should have actually set the bottom of the tin at the top of the flame every time I started using a new burner. As this would be the same for every alcohol (therefore every flame size) this would give a more reliable and constant result for each alcohol and its repeat.
Also if I was to repeat the experiment I would try to insulate this gap, even if the bottom of the tin was resting at the top of the flame. Insulation with tin foil possibly would prevent heat loss drastically.
I believe that it is these experimental flaws that were the reason for my drastically varied results. Obviously if I was to repeat this experiment I would attend to these flaw to ensure a more accurate and reliable set of results.
Flaws in how I carried out the procedure-
When reading the temperature of the thermometer in the tin I believe that the thermometer was resting on the bottom of the tin (the base). That meant that I was most likely taking the temperature more of the base of the tin that the actual water. What I should have done was to be stirring the water around with a glass rod or something similar, and also taking the thermometer reading where the bottom of the thermometer was resting in the middle of the water. To do this with my thermometer I could have used another retort stand and clamp and attached my thermometer to that. Then upon raising or lowering the tin depending on the different flame sizes I would have also raised or lowered the thermometer accordingly.
I feel that the conclusion that I have drawn is reliable as it followed my prediction. I believe that obviously to obtain more accurate and reliable results I would need to attend to the flaws I stated above. The reason for my repeat results being so far apart from each other was most probably due to the flaws stated above. I also did repeats using different burners. Probably because I lost track of which burners I had used in the classroom as others needed to use them for their experiment too. What I should have done was to take one reading with a specific alcohol. Then straight away using the same burner take a repeat. This would be a lot more reliable than using a different burner at a different time.
If I was to extend this investigation I would investigate the isomers of the alcohols I have just used in this experiment. Investigating these would allow me to then compare those results to the ones I have now. I would do exactly the same experiment obviously adding in amendments to my flaws stated above. I would then see whether an isomer of an alcohol has a different enthalpy to the alcohol itself.